Runge-Kutta fourth order method is used to solve the differential equation \( \frac{dy}{dx} =( y~-~x)\) . If the initial value y(0) = 2 and step-size is 0.1, then the value of y(0.1) is ______ (up to three decimal places)
Concept:
Consider the Differntial equation,
\(\frac{dx}{dy} = f(x_i, y_i)\) and y(xi) = yi
Then,
Fourth Order Runge-Kutta method,
yi+1 = yi + \(\frac{1}{6}\)(K1 + 2K2 + 2K3 + K4)
Where,
K1 = hf(xi, yi)
K2 = hf(xi + \(\frac{h}{2}\), yi + \(\frac{K_1}{2}\))
K3 = hf(xi + \(\frac{h}{2}\), yi + \(\frac{K_2}{2}\))
K4 = hf(xi + h, yi + K3)
h = Step size
Calculation:
Given:
\( \frac{dy}{dx} =( y~-~x)\), h = 0.1, xi = 0, yi = 2.
K1 = hf(xi, yi)
K1= 0.1f(0, 2) = 0.1(2 - 0) = 0.2
K2 = hf(xi + \(\frac{h}{2}\), yi + \(\frac{K_1}{2}\))
K2 = 0.1(0 + 0.05, 2 + 0.1) = 0.1f(0.05, 2.1)
K2 = 0.1(2.1 - 0.05) = 0.205
K3= hf(xi + \(\frac{h}{2}\), yi + \(\frac{K_2}{2}\))
K3 = 0.1f(0.05, 2 + 0.1025)
K3 = 0.1f(0.05, 2.1025)
K3 = 0.1(2.1025 - 0.05) = 0.20525
K4 = hf(xi + h, yi + K3)
K4 = 0.1f(0.1, 2 + 0.20525)
K4 = 0.1(2.20525 - 0.1) = 0.210525
Then,
K = \(\frac{1}{6}\)(K1 + 2K2 + 2K3 + K4)
= \(\frac{1}{6}\)(0.2 + 2 × 0.205 + 2 × 0.20525 + 0.210525)
= 0.1667 × 1.231025
= 0.20521
So,
y(0.1) = yi + K
= 2 + 0.20521
= 2.21
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